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mr Goodwill [35]
1 year ago
8

What might be the result of removing a top predator from a marine biome?

Chemistry
1 answer:
julsineya [31]1 year ago
4 0

Answer:

A

Explanation:

Due to the fact that no one will eat the fish, their population starts growing.

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
Hospital patients are administered oxygen from an pressurized
Nataly_w [17]

Answer:

110L

Explanation:

Boyle's Law states that P1×V1=P2×V2

Volume is indirectly proportional to Pressure so P×V is constant

P1=55atm

V1=6L

P2=3atm

V2 is to be found

P1×V1=P2×V2

6×55=3×V2

330=3×V2

Answer: V2=110L

8 0
3 years ago
1) Какое количество вещества Феррум (III) гидроксида вступает в реакцию с 15 г сульфатной кислоты?
EleoNora [17]

i dont speak russian sorry

3 0
3 years ago
How many grams of H, are needed to react with 2.75 g of N,?
Elena-2011 [213]

Answer:

0.6 grams of hydrogen are needed to react with 2.75 g of nitrogen.

Explanation:

When hydrogen and nitrogen react they form ammonia.

Chemical equation:

N₂ + 3H₂ → 2NH₃

Given mass of nitrogen = 2.75 g

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 2.75 g / 28 g/mol

Number of moles = 0.098 mol

Now we will compare the moles of nitrogen with hydrogen from balance chemical equation:

                   N₂             :          H₂

                    1               :           3

                   0.098       :        3×0.098 = 0.3 mol

Mass of hydrogen:

Mass = number of moles × molar mass

Mass = 0.3 mol × 2 g/mol

Mass = 0.6 g

6 0
3 years ago
Given the balanced equation:
Svet_ta [14]
Is there any answer choices before i get started on working out the problem
8 0
2 years ago
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